• Standardizing observations for comparisons
• The normal model
• 68-95-99.7 rule for the normal model
• Finding percentiles and areas for given observations from a normal model
• Finding observations for given percentiles or areas from a normal model

• ACT score
• SAT score
• Different scale
• Measuring similar values
• How do you compare them?

• The average score on the ACT English exam is 21.0, with a standard deviation of 4.0. The average score on the SAT verbal exam is 520 with a standard deviation of 100.
  • Albert scores a 27 on the ACT English exam
  • Alfred scores a 770 on the SAT Verbal exam
• Who has the better score? Albert or Alfred?

• \(y =\) observation of quantitative variable
• How does the value of \(y\) relate to the mean value?
• How does the value of \(y\) for this quantitative variable relate to another observation of a different quantitative variable?

\[z = \frac{y - \bar{y}}{s}\]

• \(z\) has no units (just a number)
• Puts observations on same scale
  • Mean (center) at 0
  • Standard deviation (spread) of 1
• Does not change overall shape of the distribution

• \(z =\) # of standard deviations observation is away from mean
  • Negative \(z\): observation is below mean
    • Ex: \(z=-1.5\Rightarrow\) observation is 1.5 standard devations below mean
  • Positive \(z\): observation is above mean
    • Ex: \(z=0.5\Rightarrow\) observation is 0.5 standard deviations above mean

• The average score on the ACT English exam is 21.0, with a standard deviation of 4.0. The average score on the SAT verbal exam is 520 with a standard deviation of 100.
  • Albert scores a 27 on the ACT English exam
  • Alfred scores a 770 on the SAT Verbal exam
• How do we answer who has the better score?
  • Compute z-scores and compare

• Albert \[z = \frac{y_{1} - \bar{y}}{s} = \frac{27-21}{4} = 1.5\]
• Alfred \[z = \frac{y_{2} - \bar{y}}{s} = \frac{770 - 520}{100} = 2.5\]
• Who has the better scores?

• The average score on the ACT Math exam is 20.7 with a standard deviation of 4.1. The average score on the SAT Math exam is 510 with a standard deviation of 100
  • Alberta scores a 15 on the ACT Math exam
  • Fredda scores a 340 on the SAT Math exam
• Who has the better score? Alberta or Fredda?

• Alberta \[z = \frac{y_{1} - \bar{y}}{s} = \frac{15-20.7}{4.1} = -1.39\]
• Fredda \[z = \frac{y_{2} - \bar{y}}{s} = \frac{340 - 510}{100} = -1.70\]
• Who has the better score?

• Standardizing
  • Allows you to make comparisons of observations between different variables
  • Without the distribution information, you still don't know anything about the percentile value of your observation
  • This percentile value depends on the distribution

• Different models according to different aspects of distributions
  • Shape
  • Center
  • Variability

• Shape
  • Unimodal
  • Symmetric
  • Bell-shaped
• Determined by two parameters
  • Mean \((\mu)\)
  • Standard deviation \((\sigma)\)

• No data distribution follows a normal model exactly
• Many data distributions are very close though
• How do you know?
  • Histogram
  • Normal Quantile Plot (will learn in lab)

• Mean \(\mu\)
  • Locates the center of the distribution
  • Splits the curve in half
• Standard deviation \(\sigma\)
  • Controls the variability of curve
  • Ruler of distribution
• Write as \(N(\mu, \sigma)\)

• Height of men is normally distributed
  • Mean \(\mu = 70\) inches
  • Standard deviation \(\sigma = 3\) inches
• So, the height of men is distributed \(N(70,3)\)
• What percent of men have heights less than 70 inches?
• What percent of men have heights greater than 70 inches?

• Approximately 68% of observations are within 1 \(\sigma\) of the mean \(\mu\)
• If \(\mu = 0\) and \(\sigma = 1\), approximately 68% of the observations are between -1 and 1

• Approximately 95% of observations are within \(2\sigma\) of the mean \(\mu\)
• If \(\mu = 0\) and \(\sigma = 1\), approximately 95% of the observations are between \(-2\) and \(2\)

• Approximately 99.7% of observations are within \(3\sigma\) of the mean \(\mu\)
• If \(\mu = 0\) and \(\sigma = 1\), approximately 99.7% of the observations are between \(-3\) and \(3\)

• Height of men \(\sim N(70, 3)\)
• 68-95-99.7 rule
  • 68% of men will have heights between which two values?
    • \(\mu-\sigma = 70 - 3 = 67\) inches
    • \(\mu+\sigma = 70 + 3 = 73\) inches

• Height of men \(\sim N(70,3)\)
• 68-95-99.7 rule
  • 95% of men will have heights between which two values?
    • \(\mu - 2\sigma = 70 - 2*2 = 64\) inches
    • \(\mu + 2\sigma = 70 + 2*2 = 76\) inches

• Height of men \(\sim N(70,3)\)
• 68-95-99.7 rule
  • 99.7% of men will have heights between which two values?
    • \(\mu - 3\sigma = 70 - 3*2 = 61\) inches
    • \(\mu + 3\sigma = 70 + 3*2 = 79\) inches

• Z-table
  • Connection between normal values and percentiles for standard normal model
• Standard normal model
  • Denoted by \(Z\)
  • Mean \(\mu = 0\)
  • Standard deviation \(\sigma = 1\)

• Table gives proportion of curve below a particular \(z\) score (the percentile for the value \(z\))
  • \(z\) values range from \(-3.90\) to \(3.90\) on the table
  • Row - ones and tenths place for \(z\)
  • Column - hundredths place for \(z\)

• Changes any normal model to a standard normal model \[z = \frac{y - \mu}{\sigma}\]
• \(z =\) # of standard deviations away from mean \(\mu\)
  • Negative \(z\Rightarrow\) number is below mean
  • Positive \(z\Rightarrow\) number is above mean

• The height of men follows a normal distribution with mean 70 inches and standard deviation 3 inches
  • Standardize \(y=68\) \[z = \frac{68-70}{3} = -0.67\]
  • Standardize \(y=74\) \[z = \frac{74-70}{3} = 1.33\]

• Finding areas

1. Standardize \(y\) to get \(z\)
2. Find percentile for \(z\) from Z-table
3. Use percentile to obtain needed area

• Finding normal values

1. Determine percentile from given information in problem
2. Find \(z\) value for percentile from Z-table
3. Calculate normal value \(y\) using equation \[y = \mu + z\sigma\]

• Height of men \(\sim N(70,3)\)
• What proportion of men are shorter than 66 inches?

\[z = \frac{66 - 70}{3} = -1.33\]

• Height of men \(\sim N(70,3)\)
• What proportion of men are taller than 74 inches? \[z = \frac{74 - 70}{3} = 1.33\]

• Height of men \(\sim N(70,3)\)
• What proportion of men are between 68 and 71 inches tall? \[z_{1} = \frac{68 - 70}{3} = -0.67\] \[z_{2} = \frac{71 - 70}{3} = 0.33\]

• The height of men is known to follow a normal model with mean 70 and standard deviation 3
• 20% of men are taller than what height?
  • Look up 0.8 in the table and work backwards to find \(z\)
  • \(z = 0.84\)
  • \(y = 70 + 0.84(3) = 72.52\)

• The height of men is known to follow a normal model with mean 70 and standard deviation 3
• 65% of men are shorter than what height?
  • Look up 0.65 in the table and work backwards to find \(z\)
  • \(z = 0.39\)
  • \(y = 70 + 0.39(3) = 71.17\)

• The height of men is known to follow a normal model with mean 0 and standard deviation 3
• Would it be unusual to find a man that is more than 80 inches tall? Why or why not?
  • His z-score is \(3.33\). He is more than 3 standard deviations above the mean. So yes, that is unusual
  • The probability of being 80" tall or more is 0.0004
• Would it be unusual to find a man that is less than 59 inches tall? Why or why not?
  • His z-score is -\(3.67\). He is more than 3 standard deviations below the mean. So yes, that is unusual
  • The probability of being less than or equal to 59" tall is 0.0001

• If you have an observation that is more than 3 standard deviations from the mean (z-score more than 3 or less than -3), it is considered to be an unusual occurence